Is there a reason why shared_ptr does not overload operator->*? That operator would ease the access of the pointee's member via "pointer to member":

I'd like to access them via shp->*member but that does not work, and I have to write (*shp).*member which looks awkward and unnaturally to me. --31.18.70.70 06:37, 15 December 2014 (PST)

You could also write shp.get()->*member or (&*shp)->*member if you want the correct operator to be called. Some early smart pointers overloaded ->*, but the boost implementation of shared_ptr, on which C++ is based, didn't. I see it was recently brought up on stackoverflow with no conclusive answer. --Cubbi (talk) 06:50, 15 December 2014 (PST)